Q.If $A$ and $B$ are matrices of same order, then $(AB' - BA')$ is a
(A) skew symmetric matrix
(B) null matrix
(C) symmetric matrix
(D) unit matrix

The key idea is that for any matrix $X$, the expression $X - X'$ is always skew symmetric, because $(X - X')' = X' - X = -(X - X')$.

Let $M = AB' - BA'$.

Take its transpose:

$M' = (AB' - BA')' = (AB')' - (BA')' = (B')'A' - (A')'B' = BA' - AB' = -(AB' - BA') = -M$. …

Why This Works: The Core Idea

The problem asks about the nature of the matrix $M = AB' - BA'$, where $A$ and $B$ are of the same order. The key is to check whether $M$ is symmetric ($M' = M$) or skew symmetric ($M' = -M$).

The transpose operation has a beautiful property: it reverses the order of a product. So $(AB')' = (B')' A' = BA'$. This reversal is the engine behind the entire proof.

Step-by-Step Reasoning

1. Write down the given matrix.

Let $M = AB' - BA'$. We need to find $M'$, the transpose of $M$.

2. Take the transpose of $M$.

Using the property that $(P - Q)' = P' - Q'$, we get:

3. Apply the reversal rule for transposes.

For any two matrices $X$ and $Y$, $(XY)' = Y' X'$. So:

because $(B')' = B$. Similarly:

4. Substitute back.

5. Compare $M'$ with $M$. …

Method: Determine the symmetry type of a matrix expression

Use this whenever you must decide if an expression like $AB' - BA'$ is symmetric, skew-symmetric, or neither.

Steps

Step 1: Call the expression $M$ and compute its transpose $M'$.

Symmetry is decided entirely by comparing $M'$ with $M$: $M' = M$ is symmetric, $M' = -M$ is skew-symmetric.

Step 2: Apply the transpose rules carefully.

Use the reversal law and linearity:

$(XY)' = Y'X', \qquad (X \pm Y)' = X' \pm Y', \qquad (X')' = X.$ …

Common Mistakes

Mistake 1: Writing $(AB')' = A'B'$ instead of reversing the order.

Why it's wrong: the transpose of a product reverses the factors: $(AB')' = (B')'A' = BA'$. Correct approach: always flip the order when transposing a product.

Mistake 2: Leaving a double transpose as $(B')'$ instead of simplifying to $B$.

Why it's wrong: not simplifying $(B')' = B$ breaks the cancellation that reveals $M' = -M$. Correct approach: reduce every $(X')'$ to $X$. …